Chapter 14 Partial Derivatives 14.1 Functions of Several Variables 14.2 Limits and Continuity 14.3 Partial Derivatives 14.4 Tangent Planes and Linear Approximations 14.5 The Chain Rule *14.6 Directional Derivatives and the Gradient Vector 14.7 Maximum and Minimum Values 14.8 Lagrange Multipliers

So far we have dealt with the calculus of functions of a single variable. But, in the real world, physical quantities often depend on two or more variables, so in this chapter we turn our attention to functions of several variables and extend the basic ideas of differential calculus to such functions.

14.1 Functions of Several Variables In this section we study functions of two or more variables from four points of view: Verbally (by a description in words) Numerically (by a table of values) Algebraically (by an explicit formula) Visually (by a graph or level curves)

Functions of Two Variables Definition A function of two variables is a rule that assigns to each ordered pair of real numbers (x,y) in a set D a unique real number denoted by f(x,y). The set D is the domain of f and its range is the set of values that f takes on, that is , {f(x,y)| (x,y) D}. We often write z= f(x,y) to make explicit the value taken on by f at the general point (x,y) . The variables x and y are independent variables and z is the dependent variable. y f x z f(x,y) （x,y）

Example 1 Find the domains of the following functions and evaluate f(3,2). (a) (b) Solution (a) (b)

Graphs Another way of visualizing the behavior of a function of two variables is to consider its graph. Definition If f is a function of two variables with domain D, then the graph of f is the set of all points (x,y,z) in such that z=f(x,y) and (x,y) is in D. z S y D x

Functions of Three or More Variables A function of three variables, f, is a rule that assigns to each ordered triple (x,y,z) in a domain D a unique real number denoted by f(x,y,z). A function of n variables, f, is a rule that assigns a number to an n-tuple of real number. We denoted by the set of all such n-tuples.

14.2 Limits and Continuity Definition Let f be a function of two variables whose domain D includes points arbitrarily close to (a,b). Then we say that the limit of f(x,y) as (x,y) approaches (a,b) is L and we write if for every number there is a corresponding number such that whenever and

If as along a path and as along a path , where , then does not exist. Example does not exist. Solution as along the x-axis as along the y-axis

Continuity Definition A function f of two variables is called continuous at (a,b) if We say f is continuous on D if f is continuous at every point (a,b) in D. Example

14.3 Partial Derivatives If f is a function of two variables , its partial derivatives are the functions and defined by Notations for partial derivatives If , we write

Rule for finding partial derivatives of 1.To find regard as a constant and differentiate with respect to 2.To find regard as a constant and differentiate with respect to Example If , find and . Solution

Interpretations of partial derivatives Stand for differentiate of tangent through of curve Stand for differentiate of tangent through of curve 机动 目录 上页 下页 返回 结束

Higher derivatives If then

Example Clairaut’s Theorem Suppose f is defined on a disk D that contains the point (a,b). If the functions and are both continuous on D, then

14.4 Tangent Planes and Linear Approximations Definition If , then f is differentiable at (a,b) if can be expressed in the form where and as . Theorem If the partial derivatives and exist near (a,b) and are continuous at (a,b), then f is differentiable at (a,b).

Differentials For a differentiable function of two variables, , we define the differentials and to be independent variables; that is, they can be given any values. Then the differential , also called the total differential , is defined by

14.5 The Chain Rule The Chain Rule (Case 1) Suppose that is a differentiable function of and , where and are both differentiable functions of . Then is a differentiable function of and or

Example Let z = excosy, x = sint and y = t2. Find Solution We find that The Chain Rule (Case 2) Suppose that is a differentiable function of and , where and are differentiable functions of and .Then

Example Let z = xy, x = 3u2 + v2, and y = 4u +2v. Find and Solution

Implicit Differentiation We suppose that an equation of the form defines implicitly as a differentiable function of , that is, , where for all in the domain of f. If F is differentiable. Then

We suppose that an equation of the form defines implicitly as a differentiable function of and , that is, , where for all in the domain of f. If F is differentiable. Then

Example 1 Find if Solution Let then Example 2 Find and if

14.7 Maximum and Minimum Values Definition A function of two variables has a local maximum at (a,b) if when (x,y) is near (a,b). [ This means that for all points (x,y) in some disk with center (a,b).] The number f(a,b) is called a local maximum value. If when (x,y) is near (a,b), then f(a,b) is called a local minimum value. If the inequalities hold for all points (x,y) in the domain of f, then f has an absolute maxi- mum (or absolute minimum) at (a,b).

Theorem If f has a local maximum or minimum at (a,b) and the first-order partial derivatives of f exist there, then and . A point (a,b) is called a critical point (or stationary point ) of f if and , or one of partial Derivatives does not exist. Theorm says that if f has a local maximum or mini- mum at (a,b), then (a,b) is a critical point of f . However, not all critical points give rise to maximum or minimum.

Second derivatives test Suppose the second partial derivatives of f are continuous on a disk with center (a,b), and suppose that and [that is, (a,b) is a critical point of f ]. Let (a) If and , then f(a,b) is a local minimum. (b) If and , then f(a,b) is a local maximum. (c) If ,then f(a,b) is not a local minimum or local

Note 1 In case (c) the point (a,b) is called a saddle point of f and the graph of f crosses its tangent plane at (a,b). Note 2 If D=0, the test gives no information: f could have a local minimum or local maximum at (a,b), or (a,b) could be a saddle point of f . Note 3 We often use the following symbols:

Example Let f(x, y) = x2 – y2. Show that the origin is the only critical point but that f has no extreme value at the origin. Solution The partial derivatives of f exist at every point in the domain of f, and we have that f’x(x, y) = 2x, f’y(x, y) = -2y, so that (0, 0) is the only critical point of f. However, f(0, 0) = 0 is no a extreme value of f, because f(x, 0) = x2 > 0 for all x≠0 and f(0, y) = -y2 < 0 for all y ≠0.

Example Given f(x, y) = x3 +y3 –3xy, find the extreme values of function f. Solution f has continuous second partial derivatives so the critical points of f are those at which f’x and f’y are 0. Since f’x(x, y) = 3x2 –3y and f’y(x, y) = 3y2 –3x, Solving for x, y from 3x2 –3y = 0, 3y2 –3x = 0, we obtain the critical points of f, which are (1, 1) and (0, 0). For (1, 1) we have A = 6, B = -3, C = 6, and D=27> 0. Since A>0, we know that f(1, 1) = -1 is a local minimum. For (0, 0) we have A = 0, B = -3, C = 0, and D= -9< 0, hence f has no extreme value at (0, 0).

Absolute Maximum and Minimum Values Extreme value theorem for functions of two variables If f is continuous on a closed, bounded set D in , then f attains an absolute maximum value and an absolute minimum value at some points and in D. To find the absolute maximum and minimum values of a continuous function of f on a closed, bounded set D : 1. Find the values of f at the critical points of f in D. 2. Find the extreme values of f on the boundary of D. 3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.

Example Find the absolute maximum and minimum values of the function on Solution First find the critical points. These occur when So the only critical point is (1,1), and the value is f(1,1)=1. In step 2 we look at the values of f on the boundary of D, which consists of the four line segments of D. We attain that f(0,0)=0 and f(3,2)=1 are minimum values of f on the boundary of D , we also attain that f(3,0)=9 and f(0,2)=4 are maximum values of f on the boundary of D . In step 3 we compare these values and conclude that the absolute maximum value of f on D is f(3,0)=9 and the absolute minimum value is f(0,0)=0.

14.8 Lagrange Multipliers Conditional extreme values and Lagrange multipliers In previous discussion on extreme values of functions of two variables x and y, the variables x and y are independent to each other, i.e., they are not constrained by any side condition on the values of x and y. However, we may encounter problems of finding the extreme values of function f(x, y) subject to a side condition, also called a constraint, of the form g(x, y) = 0. Such extreme values are called conditional extreme values.

We will describe a general method of finding the extreme values of a function subject to a constraint, which is called the Lagrange multipliers method, and the number is called a Lagrange multiplier. The method of determining extreme values of function f(x, y) subject to constraint g(x, y) = 0 by means of Lagrange multipliers proceed as follows.

1. Form a function F by multiplying g(x, y) with the Lagrange multiplier and then adding to f(x, y), that is, 2. Find the partial derivatives of F, F’x and F’y , and then solve the equations 3. Calculate the value of f at each point (x, y) that is obtained in step 2.

Then, subject to the constraint g(x, y) = 0, if f(x, y) has a (conditional) maximum value, it will be the largest of the values computed; if f(x, y) has a (conditional) minimum value, it will be the smallest of the values computed. Example 1 Find the extreme values of f(x, y) = x2 + 4y3 subject to the constraint x2 + 2y2 =1. Solution Let Then Solving

we obtain as the solutions we obtain as the solutions. Thus the only possible extreme values of f occur at points in the xy plane:

Since we conclude that the maximum value of f(x, y) on the ellipse is , which occurs at point , and the minimum value of f(x,y) on the ellipse is , which occurs at point .

Example 2 A rectangular box without a lid is to be made from 12 m2 of cardboard. Find the maximum volume of such a box. Solution Let x, y, and z be the length, width, and height, respectively, of the box in meters. We wish to maximize V= xyz subject to the constraint g(x,y,z)=2xz+2yz+xy-12=0 Let Solving the equations

we have x=y=2 and z=1. Then V=xyz=4, so the maximum volume of the box is 4 m3.